如何在Node.js中获取脚本的路径?

我知道有流程。Cwd,但它只引用调用脚本的目录,而不是脚本本身。例如,假设我在/home/kyle/目录下,然后运行以下命令:

node /home/kyle/some/dir/file.js

如果我调用process.cwd(),我会得到/home/kyle/,而不是/home/kyle/some/dir/。有办法得到那个目录吗?


当前回答

当涉及到主脚本时,它非常简单:

process.argv[1]

来自Node.js文档:

process.argv 包含命令行参数的数组。第一个元素是“node”,第二个元素是JavaScript文件的路径。下一个元素将是任何额外的命令行参数。

如果你需要知道模块文件的路径,那么使用__filename。

其他回答

I know this is pretty old, and the original question I was responding to is marked as duplicate and directed here, but I ran into an issue trying to get jasmine-reporters to work and didn't like the idea that I had to downgrade in order for it to work. I found out that jasmine-reporters wasn't resolving the savePath correctly and was actually putting the reports folder output in jasmine-reporters directory instead of the root directory of where I ran gulp. In order to make this work correctly I ended up using process.env.INIT_CWD to get the initial Current Working Directory which should be the directory where you ran gulp. Hope this helps someone.

var reporters = require('jasmine-reporters');
var junitReporter = new reporters.JUnitXmlReporter({
  savePath: process.env.INIT_CWD + '/report/e2e/',
  consolidateAll: true,
  captureStdout: true
 });

当涉及到主脚本时,它非常简单:

process.argv[1]

来自Node.js文档:

process.argv 包含命令行参数的数组。第一个元素是“node”,第二个元素是JavaScript文件的路径。下一个元素将是任何额外的命令行参数。

如果你需要知道模块文件的路径,那么使用__filename。

如果你想在shell脚本中使用类似$0的东西,试试这个:

var path = require('path');

var command = getCurrentScriptPath();

console.log(`Usage: ${command} <foo> <bar>`);

function getCurrentScriptPath () {
    // Relative path from current working directory to the location of this script
    var pathToScript = path.relative(process.cwd(), __filename);

    // Check if current working dir is the same as the script
    if (process.cwd() === __dirname) {
        // E.g. "./foobar.js"
        return '.' + path.sep + pathToScript;
    } else {
        // E.g. "foo/bar/baz.js"
        return pathToScript;
    }
}

Node.js 10支持ECMAScript模块,其中__dirname和__filename不再可用。

然后,要获得当前ES模块的路径,必须使用:

import { fileURLToPath } from 'url';

const __filename = fileURLToPath(import.meta.url);

对于包含当前模块的目录:

import { dirname } from 'path';
import { fileURLToPath } from 'url';

const __dirname = dirname(fileURLToPath(import.meta.url));

使用__dirname ! !

__dirname

当前模块的目录名。这与__filename的path.dirname()相同。

例如:在/Users/mjr中运行node Example .js

console.log(__dirname);
// Prints: /Users/mjr
console.log(path.dirname(__filename));
// Prints: /Users/mjr

https://nodejs.org/api/modules.html#modules_dirname

对于esmodule,你会想要使用: import.meta.url